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If $\sin \theta + \cos \theta = p$ and $\sec \theta + \text{cosec } \theta = q$, then prove that $q(p^2 - 1) = 2p$.
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$\sin \theta + \cos \theta = p$, $\sec \theta + \text{cosec } \theta = q$
LHS = $q(p^2 - 1)$
$= (\sec \theta + \text{cosec } \theta)[(\sin \theta + \cos \theta)^2 - 1]$
$= (\frac{1}{\cos \theta} + \frac{1}{\sin \theta})[\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta - 1]$
$= (\frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta}) [1 + 2 \sin \theta \cos \theta - 1]$
$= (\frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta}) (2 \sin \theta \cos \theta)$
$= 2(\sin \theta + \cos \theta)$
$= 2p = RHS$
LHS = $q(p^2 - 1)$
$= (\sec \theta + \text{cosec } \theta)[(\sin \theta + \cos \theta)^2 - 1]$
$= (\frac{1}{\cos \theta} + \frac{1}{\sin \theta})[\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta - 1]$
$= (\frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta}) [1 + 2 \sin \theta \cos \theta - 1]$
$= (\frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta}) (2 \sin \theta \cos \theta)$
$= 2(\sin \theta + \cos \theta)$
$= 2p = RHS$